Question

Please help, I need input on if I did this correctly and you just substitute.

Answer 1

The value of h(t) when $t=\frac{15}{32}$ is 10.02.

Solution:

Given function $h(t)=-16t^2+15t+6.5$

To find the value of h(t) when $t=\frac{15}{32}$:

$h(t)=-16t^2+15t+6.5$

Substitute $t=\frac{15}{32}$ in the given function.

$h\left(\frac{15}{32} \right)=-16\left(\frac{15}{32} \right)^2+15\left(\frac{15}{32} \right)+6.5$

            $=-16\left(\frac{225}{1024} \right)+15\left(\frac{15}{32} \right)+6.5$

Now multiply the common terms into inside the bracket.

           $=-\left(\frac{3600}{1024} \right)+\left(\frac{225}{32} \right)+6.5$

Now, in the first term, the numerator and denominator both have common factor 16. So reduce the first term into the lowest term.

          $=-\left(\frac{225}{64} \right)+\left(\frac{225}{32} \right)+6.5$

To make the denominator same, take LCM of the denominators.

LCM of 64 and 32 = 64

        $=-\left(\frac{225}{64} \right)+\left(\frac{225\times2}{32\times2} \right)+6.5\times\frac{64}{64}$

        $=-\frac{225}{64} +\frac{450}{64}+\frac{416}{64}$

        $=\frac{-225+450+416}{64}$

       $=\frac{641}{64}$

       = 10.02

$h\left(\frac{15}{32} \right)=10.02$

Hence the value of h(t) when $t=\frac{15}{32}$ is 10.02.